Since the Hadamard product is commutative [1]}, we can reorder the products in previous equation. The Hadamard product
\(\left( u \cdot \tilde{d}^{T}\right) \circ \left( u\cdot d^{T}\right)\)  equals all-one matrix \(J\) . Similarly, the product \(\left( \tilde{d} \cdot u^{T}\right)\circ \left( d\cdot u^{T}\right) = J\)  . We further transform the Hadamard product of \(\left( A\circ A^{2}+A\right) \)  and the outer
products \(\left( u \cdot \tilde{d}^{T}\right) \)  and \(\left( \tilde{d} \cdot u^{T}\right) \)  into product with diagonal matrices \(\Delta ^{-1}\cdot \left( A\circ A^{2}+A\right) \cdot \Delta ^{-1}\) . Thus, equation (REF ) transforms to (REF ).
Substituting (REF ) into (REF ) yields
\(x[k+1] - x[k] = \bigg (\left( u\cdot x^{T}[k] -x[k]\cdot u^{T}\right) \circ W\bigg )\cdot u.\) 
